Simplification of Expression

By: Ali Mustafa

How to simplify the Boolean Function

• Algebric method

• K-Map method (Karnaugh)

• Quine – Mc Cluskey method

• VEM (Variable entered mapping)

Algebraic Forms( A way to represent any function)

Sum of Products (SOP)

Product of Sums (POS)

Sum of Products (SOP)

• Switching functions formed by SUMMING (ORing) PRODUCT (ANDed) terms.

Sum-of-products(SOP)

• Product terms are known as minterm

• Output = 1 , for SOP

Sum-of-products (cont.)• ANDed product–input combination for which output

is true

• Each variable appears exactly once, in true or inverted form (but not both)

Sum-of-products

Example : Simplify the following three variable boolean expression Y = ∑ m (2,4,6)

Solution: y = m2 +m4 + m6

=A’BC’+AB’C’+ABC’=A’BC’+AC’(B’+B)

=A’BC’+AC’= C’(A’B + A)

Hint =(A’+ A = 1) = C’(A’+ A) (B + A)

= C’(B + A)

How to build a circuit from the (SOP)function?

F = AB'C + A'B’C + AB'C + ABC' + ABC

Answer: SOP AND/OR Two-level Implementation

Product of Sums (POS)

• Switching functions formed by taking the PRODUCT (ANDing) of SUM (ORed) terms.

Product-of-sums

• Sum terms are known as Maxterm

• Output = 0 , POS

Product-of-sums (cont’d)• Sum term (or maxterm)

• ORed sum of literals –input combination for which output is false

• Each variable appears exactly once, in true or inverted form (but not both)

Product-of-sums (Self Task)

• Example : Simplify the following three variable boolean expression Y = ∏ M (1,3,5)

?

How to build a circuit from (POS) function?

F(A, B, C) = (A + B + C) (A + B' + C) (A + B’ + C’)

Answer

POS: OR/AND Two-level Implementation

SOP and POS Represent the same function

POS vs. SOP

• Any expression can be written either way

• Can convert from one to another using theorems

• Sometimes SOP looks simpler

AB + CD = ( A + C )( B + C )( A + D )( B + D )

• Sometimes POS looks simpler

(A + B)(C + D) = BD + AD + BC + AC

• SOP will be most commonly used

Standard or canonical SOP & POS forms

Sr # Logical Expression Type of Expression

1 Y = AB + ABC’ + A’BC NON STANDARD

2 Y = AB + A’B + A’B’ STANDARD

3 Y = (A’ + B)(A + B)(A + B’) STANDARD

4 Y = (A’ + B)(A + B +C) NON STANDARD

Conversion Procedure for Standard SOP

1. For Each term ,we find the missing literal

2. Then, we AND term with the term formed by ORing the missing literal and its compliment

Y = AB + AC’ + BC

Conversion Procedure for Standard SOPExample: Convert the expression into the standard SOP form

Y = AB + AC’ + BC

Solution: Given expression Y = AB + AC’ + BC

Step 1: Find the missing literals

Y = AB + AC’ + BC

Step 2: AND each term with (Missing literal + its compliment)

Y = AB(C + C’) + AC’(B + B’) + BC(A +A’)

Simplify the expression to get the standard SOP form

Y = ABC + ABC’ + ABC’+ AB’C’ + ABC + A’BC

Y = ABC + ABC’+ AB’C’ + A’BC(Each term consists of all literal)

C B A

A + A = A

Conversion Procedure for Standard POS• Example: Convert the expression into the standard SOP form

Y = ( A+B )( A+C )( B +C’)

Step 1: Find the missing literals

Step 2: OR each term with (Missing literal + its compliment)

Y = ( A + B + CC’ )( A+ BB’ +C )( AA’ + B +C’ )

Y =( A + B + C )( A + B + C’ ) ( A+ B +C )( A+ B’ +C ) ( A + B +C’ )( A’ + B +C’ )

Y =( A + B + C )( A + B + C’ )( A+ B’ +C )( A’ + B + C’ )

SELF TASKS

• Convert the following expressions into standard SOP & POS forms

1. Y = AB + AC + BC

2. Y = (A + B) (B’ + C )

3. Y = A + BC + ABC

Boolean Expression from a Logic Circuit

• To derive the Boolean expression for a given logic circuit, begin at the left-most inputs and work toward the final output, writing the expression for each gate.

Example